1. Tardigrade
  2. Question
  3. Mathematics
  4. Statement-1: The variance of first n even natural numbers is (n2-1/4) Statement-2: The sum of first n natural numbers is (n(n+1)/2) and the sum of squares of first n natural numbers is (n(n+1)(2n+1)/6)

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. Statement-1 : The variance of first $n$ even natural numbers is $\frac{n^{2}-1}{4}$
Statement-2 : The sum of first $n$ natural numbers is $\frac{n\left(n+1\right)}{2}$ and the sum of squares of first $n$ natural numbers is $\frac{n\left(n+1\right)\left(2n+1\right)}{6}$

AIEEEAIEEE 2009Statistics

Solution:

Statement-2 is true
Statement-1:
Sum of n even natural numbers $= n (n + 1)$
Mean $\left(\bar{x}\right)=\frac{n\left(n+1\right)}{n}=n+1$
Variance $=\left[\frac{1}{n}\sum\left(x_{i}\right)^{2}\right]-\left(\bar{x}\right)^{2}=\frac{1}{n}\left[2^{2}+4^{2}+.....+\left(2n\right)^{2}\right]-\left(n+1\right)^{2}$
$=\frac{1}{n}2^{2}\left[1^{2}+2^{2}+.....+n^{2}\right]-\left(n+1\right)^{2}=\frac{4}{n} \frac{n\left(n+1\right)\left(2n+1\right)}{6}-\left(n+1\right)^{2}$
$=\frac{\left(n+1\right)\left[2\left(2n+1\right)-3\left(n+1\right)\right]}{3}= \frac{\left(n+1\right)\left[4n+2-3n-3\right]}{3}=\frac{\left(n+1\right)\left(n-1\right)}{3}=\frac{n^{2}-1}{3}$
$∴$ Statement 1 is false.